A tortoise and a hare are competing in a 1200-meter race. The arrogant hare decides to let the tortoise have a 570-meter head start. When the start gun is fired the hare begins running at a constant speed of 8 meters per second and the tortoise begins crawling at a constant speed of 6 meters per second.

Let 
t
 represent the number of seconds that have elapsed since the start gun was fired.

Write an expression in terms of 
t
 that represents the hare's distance from the starting line (in meters).

8t
    
8
t
 syntax ok

Write an expression in terms of 
t
 that represents the tortoise's distance from the starting line (in meters).

6t
    
6
t
 syntax ok

Write an expression in terms of 
t
 that represents the number of meters the tortoise is ahead of the hare.

   

Who finishes the race first?

Let's break down the problem step by step to form the required expressions and then determine who finishes the race first.

### 1. Hare's distance from the starting line:
The hare runs at a constant speed of \( 8 \, \text{m/s} \). So, the hare's distance from the starting line at any time \( t \) seconds is:

\[
\text{Hare's distance} = 8t
\]

### 2. Tortoise's distance from the starting line:
The tortoise starts with a head start of 570 meters and moves at a constant speed of \( 6 \, \text{m/s} \). The tortoise's distance from the starting line at any time \( t \) seconds is:

\[
\text{Tortoise's distance} = 570 + 6t
\]

### 3. The number of meters the tortoise is ahead of the hare:
The tortoise's lead over the hare is the difference between the tortoise's and the hare's distances from the starting line. This can be written as:

\[
\text{Tortoise's lead} = (570 + 6t) - 8t = 570 - 2t
\]

### 4. Who finishes the race first?
To determine who finishes the race first, we need to calculate the time it takes for both the hare and the tortoise to complete the 1200-meter race.

- For the hare:
  \[
  8t = 1200
  \]
  Solving for \( t \):
  \[
  t = \frac{1200}{8} = 150 \, \text{seconds}
  \]

- For the tortoise:
  \[
  570 + 6t = 1200
  \]
  Solving for \( t \):
  \[
  6t = 1200 - 570 = 630
  \]
  \[
  t = \frac{630}{6} = 105 \, \text{seconds}
  \]

### Conclusion:
The tortoise finishes the race first, in 105 seconds, while the hare finishes in 150 seconds.

---

Would you like more details on any part of the solution? 

Here are 5 related questions to explore further:
1. How much distance did the hare cover by the time the tortoise finished the race?
2. At what time \( t \) will the hare catch up to the tortoise during the race?
3. What would happen if the hare ran at 10 m/s instead of 8 m/s?
4. How far ahead is the tortoise when it finishes the race?
5. If the tortoise's speed increased to 7 m/s, how would that affect the outcome?

**Tip:** When solving motion problems, setting up equations based on distance, speed, and time relationships is the key to finding the solution.

A tortoise and a hare are competing in a 1200-meter race. The arrogant hare decides to let the tortoise have a 570-meter head start. When the start gun is fired the hare begins running at a constant speed of 8 meters per second and the tortoise begins crawling at a constant speed of 6 meters per second. Let t represent the number of seconds that have elapsed since the start gun was fired. Write an expression in terms of t that represents the hare's distance from the starting line (in meters). Write an expression in terms of t that represents the tortoise's distance from the starting line (in meters). Write an expression in terms of t that represents the number of meters the tortoise is ahead of the hare. Who finishes the race first?

Explore the famous race between the tortoise and the hare using distance-time equations. Learn how to write linear expressions for their distances and determine who finishes the 1200-meter race first. This problem covers key mathematical concepts such as linear equations and speed-time relationships, suitable for students in Grades 7-9.

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